高考數(shù)學(xué)復(fù)習(xí)核心專題突破(三)　微專題5　數(shù)列求和（導(dǎo)學(xué)案）

核心專題突破(三)微專題5數(shù)列求和【課程標(biāo)準(zhǔn)】1.熟練掌握等差、等比數(shù)列的前n項(xiàng)和公式.2.掌握非等差數(shù)列、非等比數(shù)列求和的幾種常見(jiàn)方法.【題型一】分組轉(zhuǎn)化與并項(xiàng)求和[典例1](1)數(shù)列112,314,518,7116,…,(2n-1)+12n的前n項(xiàng)和SA.n2+1-12n B.2n2-nC.n2+1-12n?2 D.n2-解析：選A.方法一:Sn=1+3+5+…+2n?1+12+12=n1+2n?12+121?方法二:當(dāng)n=1,2時(shí),代入選項(xiàng)可得只有A正確.(2)已知數(shù)列{an}的前n項(xiàng)和為Sn=1-5+9-13+17-21+…+(-1)n+1(4n-3),則S15-S31的值是 (A.-30 B.-32 C.-12 D.12解析：選B.因?yàn)镾n=(1-5)+(9-13)+(17-21)+…+(-1)n+1(4所以S15=-4×7+4×15-3=29,S31=-4×15+4×31-3=61,所以S15-S31=29-61=-32.(3)已知數(shù)列an的前n項(xiàng)和為Sn,且a1=1,an+1+an=3n+1,則S101= (A.7701 B.7449 C.15401 D.14897解析：選A.當(dāng)n為偶數(shù)時(shí),Sn+1=a1+a2+a3=1+3×2+4+…+n+n2=1+3×n2n+22+n2所以S101=34×1002+2×100+1=7701(4)若an=2n+1,且bn=an,n為奇數(shù)n,n為偶數(shù),則數(shù)列bn的前解析：因?yàn)閎n=an,T2n=(2+4+6+…+2n)+(4+16+64+…+22n)=n(2+2n=4n+1?43+n答案:4n+1?43+【一題多變】[變式1]本例(2)中條件不變,若Sn=45,求n的值.解析：因?yàn)閍2m+a2m-1=-4(m∈N),且Sn=45,所以n為奇數(shù),設(shè)n=2k+1(k∈N),則Sn=-4k+4(2k+1)-3=45,解得k=11,故n=23.[變式2]本例(3)中條件不變,求S100及a101.解析：由an+1+an=3n+1可知S100=(a1+a2)+(a3+a4)+…+(a99+a100)=(3×1+1)+(3×3+1)+…+(3×99+1)=3(1+3+5+…+99)+50=3×50(1+99)2+50=7550由S101=7701可知,a101=7701-7550=151.【方法提煉】分組轉(zhuǎn)化法與并項(xiàng)法求和的常見(jiàn)類型(1)若an=bn±cn,且{bn},{cn}為等差或等比數(shù)列,可采用分組求和法求{an}的前n項(xiàng)和;(2)通項(xiàng)公式為an=bn,n為奇數(shù),cn,n為偶數(shù)的數(shù)列,其中數(shù)列{(3)若一個(gè)數(shù)列的相鄰的部分項(xiàng)的和是一個(gè)常數(shù)(或者是有規(guī)律的常數(shù)),常用并項(xiàng)法求和.【對(duì)點(diǎn)訓(xùn)練】1.已知數(shù)列an的通項(xiàng)公式為an=ncos(n-1)π,Sn為數(shù)列的前n項(xiàng)和,則S2023= (A.1009 B.1010 C.1011 D.1012【解題提示】將an=ncos(n-1)π化為an=n×?1n?1,解析：選D.因?yàn)楫?dāng)n為奇數(shù)時(shí)cos(n-1)π=1,n為偶數(shù)時(shí)cos(n-1)π=-1,所以cos(n-1)π=?1n所以an=ncos(n-1)π=n×?1nS2023=(1-2)+(3-4)+…+(2021-2022)+2023=-1011+2023=1012.2.在數(shù)列an中,a2=2,an+2=an+1-3an,Sn為an的前n項(xiàng)和,則S2022-S2021+3S2020的值為解析：因?yàn)閍2=2,an+2=an+1-3an,所以S2022-S2021+3S2020=a1+a2?a1+a3?a2+3答案:23.(2023·商洛模擬)已知等差數(shù)列{an}的前n項(xiàng)和為Sn,a5=9,S5=25.(1)求數(shù)列{an}的通項(xiàng)公式及Sn;(2)設(shè)bn=(-1)nSn,求數(shù)列{bn}的前n項(xiàng)和Tn.解析：(1)設(shè)數(shù)列{an}的公差為d,由S5=5a3=25得a3=a1+2d=5,又a5=9=a1+4d,所以d=2,a1=1,所以an=2n-1,Sn=n(1+2n?1)2(2)結(jié)合(1)知bn=(-1)nn2,當(dāng)n為偶數(shù)時(shí),Tn=(b1+b2)+(b3+b4)+(b5+b6)+…+(bn?1+=(-12+22)+(-32+42)+(-52+62)+…+[-(n-1)2+n2]=(2-1)(2+1)+(4-3)(4+3)+(6-5)(6+5)+…+[n-(n-1)][n+(n-1)]=1+2+3+…+n=n(當(dāng)n為奇數(shù)時(shí),n-1為偶數(shù),Tn=Tn?1+(-1)n·n2=(n?1)n綜上可知,Tn=(?1)【題型二】裂項(xiàng)法求和[典例2](1)在數(shù)列an中,an=1n+1+2n+1+…+nn+1n∈N+,bn=1anA.4nn+1 B.2nn+1 C.n解析：選A.因?yàn)閍n=1n+1+2n+1+…+nn+1=所以bn=1anan+1所以Sn=41?=41?1n+1(2)數(shù)列12n+1+2n?1A.4043?12 BC.4043-1 D.4045-1解析：選B.1=2=2n+1?2n?12,記12n+1123?(3)已知數(shù)列an滿足an=2n-1,則數(shù)列2nanan+1A.131 B.163 C.3031 解析：選D.因?yàn)閍n=2n-1,所以an+1=2n+1-1.所以2na=2n+1?1?2所以2nanan+1的前5=121?1-126(4)(2023·廣州模擬)若數(shù)列an滿足an=(-1)n-1·1n+1n+1,則anA.12022 B.12023 C.20212022 解析：選D.當(dāng)n為奇數(shù)時(shí),an=1n+1當(dāng)n為偶數(shù)時(shí),an=-1n所以S2022=(1+12)-(12+13)+(13+14)-…+(12021+12022=1-12023=2022【一題多變】[變式]本例(1)中,數(shù)列{an}的通項(xiàng)不變,若bn=1anan+2,求數(shù)列解析：因?yàn)閍n=n2,所以bn=1a=2(1n-1n+2所以b1+b2+…+bn=2(1-13+12-14+13-15=2(1+12-1n+1-=3-4n+6(n+1)(n+2)【方法提煉】破解裂項(xiàng)相消求和的關(guān)鍵點(diǎn)(1)定通項(xiàng):根據(jù)已知條件求出數(shù)列的通項(xiàng)公式.(2)巧裂項(xiàng):根據(jù)通項(xiàng)公式的特征進(jìn)行準(zhǔn)確裂項(xiàng),把數(shù)列的每一項(xiàng),表示為兩項(xiàng)之差的形式.(3)消項(xiàng)求和:通過(guò)累加抵消掉中間的項(xiàng),達(dá)到消項(xiàng)的目的,準(zhǔn)確求和.提醒使用裂項(xiàng)相消法求和,要注意正負(fù)項(xiàng)相消時(shí),消去了哪些項(xiàng),保留了哪些項(xiàng),切不可漏寫未被消去的項(xiàng).【對(duì)點(diǎn)訓(xùn)練】1.在數(shù)列{an}中,an=1n(n+1),若{an}的前n項(xiàng)和為20222023,解析：an=1n(n+1)=所以Sn=1-12+12-13+…+1n-1n+1=1-所以n=2022.答案:20222.(2023·咸寧模擬)設(shè){an}是各項(xiàng)都為正數(shù)的單調(diào)遞增數(shù)列,已知a1=4,且an滿足關(guān)系式:an+1+an=4+2an+1an,n(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=1an?1,求數(shù)列{bn}的前n項(xiàng)和解析：(1)因?yàn)閍n+1+an=4+2an+1an,n所以an+1+an-2an即(an+1-an又{an}是各項(xiàng)為正數(shù)的單調(diào)遞增數(shù)列,所以an+1-an=2,所以{an}是首項(xiàng)為2,公差為2的等差數(shù)列所以an=2+2(n-1)=2n,所以an=4n2(2)bn=1an?1==12(12n?1-1所以Sn=b1+b2+…+bn=12(1-13)+12(13-15)+…+12(12n?1-12n+1)=【加練備選】1.設(shè)數(shù)列n22n?12n+1的前n項(xiàng)和為SA.25<S100<25.5 B.25.5<S100<26C.26<S100<27 D.27<S100<27.5解析：選A.由n=14·4n24n2=14(1+1=14+18(12n?1-Sn=n4+18(1-13+13-15+…+12n?1-12n+1)=n4+所以S100=100×1012(2×100+1)2.已知數(shù)列an滿足an>0,an+1an=nan2+n?1,記數(shù)列an的前n項(xiàng)和為SnA.n B.n+1C.n2 D.n【解題提示】將an+1an=nan2+n?1變?yōu)閚an+1=an2+解析：選A.因?yàn)閍n>0,an+1a所以nan+1=an2+所以an=nan+1又Sn=a1+a2+…+an=(1a2-0a1)+(2a3-1a=nan+1-0a故Snan+1=n.3.(2023·長(zhǎng)沙模擬)已知等差數(shù)列an和正項(xiàng)等比數(shù)列bn滿足a1=b1=2,a7=b3=2a3,則數(shù)列(an2【解題提示】利用等差、等比數(shù)列基本量公式計(jì)算出公差和公比,求得an=n+1,bn=2n,因此得(an2-2)bn=n2結(jié)合n2+2n-1=2n2-(n-1)2裂項(xiàng)求解.解析：設(shè)公差和公比分別為d,qq>0由a1=b1=2,a7=b3=2a3得2+6d=2q2=22+2d解得d=1,q=2,因此an=n+1,bn=2n.所以(an2-2)bn=n2n2+2n?12n=2n2·2n-n?12=n2·2n+1-n?12·2設(shè)(an2?2)bn的前n項(xiàng)和為S(12×22-02×21)+(22×23-12×22)+…+n=n2·2n+1.答案:n2·2n+1【題型三】錯(cuò)位相減法求和[典例3](2020·全國(guó)Ⅰ卷)設(shè){an}是公比不為1的等比數(shù)列,a1為a2,a3的等差中項(xiàng).(1)求{an}的公比;(2)若a1=1,求數(shù)列{nan}的前n項(xiàng)和.解析：(1)設(shè){an}的公比為q,由題設(shè)得2a1=a2+a3,即2a1=a1q+a1q2.因?yàn)閍1≠0,所以q2+q-2=0,解得q=1(舍去)或q=-2.故{an}的公比為-2.(2)記Sn為{nan}的前n項(xiàng)和.由(1)及題設(shè)可得,an=(-2)n-1.所以Sn=1+2×(-2)+…+n×(-2)n-1,-2Sn=-2+2×(-2)2+…+(n-1)×(-2)n-1+n×(-2)n.可得3Sn=1+(-2)+(-2)2+…+(-2)n-1-n×(-2)n=1?(?2)n3-n所以Sn=19-(3【方法提煉】錯(cuò)位相減法求和的解題策略(1)巧分拆,即將數(shù)列的通項(xiàng)公式分拆為等差數(shù)列與等比數(shù)列積的形式,并求出公差和公比.(2)構(gòu)差式,即寫出Sn的表達(dá)式,再乘公比或除以公比,然后將兩式相減.(3)后求和,根據(jù)差式的特征準(zhǔn)確進(jìn)行求和.提醒錯(cuò)位相減法求和時(shí),應(yīng)注意:①在寫出“Sn”與“qSn”的表達(dá)式時(shí)應(yīng)特別注意將兩式“錯(cuò)項(xiàng)對(duì)齊”,以便于下一步準(zhǔn)確地寫出“Sn-qSn”的表達(dá)式.②應(yīng)用等比數(shù)列求和公式必須注意公比q是否等于1,如果q=1,應(yīng)用公式Sn=na1.【對(duì)點(diǎn)訓(xùn)練】(2023·衡水模擬)已知數(shù)列an的前n項(xiàng)和為Sn,且a2=6,an+1=2S(1)證明:an為等比數(shù)列,并求an(2)求數(shù)列nan的前n項(xiàng)和T解析：(1)因?yàn)閍n+1=2Sn所以an=2Sn?1+1故an+1-an=2Sn?Sn?1=2an,又a2=2S1+1=2a故a1=2,即a2a1=3,因此an+1a故an是以2為首項(xiàng),3為公比的等比數(shù)列因此an=2×3n-1(n∈N*).(2)因?yàn)門n=2×1+2×2×3+2×3×32+…+2n×3n-1①,故3Tn=2×1×3+2×2×32+…+2(n-1)×3n-1+2n×3n②,①-②,得-2Tn=2+2×3+2×32+…+2×3=2+2×33n?1?1=-1+(1-2n)×3n,即Tn=(2n【備選題型】倒序相加法求和與遞推關(guān)系式中含(-1)n的數(shù)列求和[典例]已知數(shù)列an滿足an+2+?1nan=3,a1=1,a2=2,數(shù)列an的前n項(xiàng)和為Sn,則S30= A.351 B.353 C.531 D.533【解題提示】由于通項(xiàng)公式中含(-1)n,因此需討論n的奇偶性,當(dāng)n為奇數(shù)時(shí),可得an+2-an=3為等差數(shù)列;當(dāng)n為偶數(shù)時(shí),可得an+2+an=3,分組求和.解析：選B.由an+2+?1nan=3可知當(dāng)n為奇數(shù)時(shí)有an+2-an=3,即有a3-a1=3,a5-a3=3,…,a2n+1-a2n-1=3.令bn=a2n-1,故bn+1-bn=3,所以數(shù)列bn是首項(xiàng)為1,公差為3的等差數(shù)列故bn=3n-2;當(dāng)n為偶數(shù)時(shí)有an+2+an=3,即a4+a2=3,a6+a4=3,…,a2n+2+a2n=3.于是S30=a1+=b1+b2+…+b15+[=1+432×15+2+7×3=330+23=353【方法提煉】(1)倒序相加法:如果一個(gè)數(shù)列的前n項(xiàng)中,距首末兩項(xiàng)“等距離”的兩項(xiàng)之和都相等,則可使用倒序相加法求數(shù)列的前n項(xiàng)和.此類問(wèn)題常與函數(shù)結(jié)合在一起命題;(2)數(shù)列的遞推關(guān)系式中含(-1)n的求和問(wèn)題,常需要根據(jù)n的奇偶性的特征,構(gòu)造數(shù)列分組轉(zhuǎn)化或?qū)ふ乙?guī)律后求解.【對(duì)點(diǎn)訓(xùn)練】1.(2022·岳陽(yáng)模擬)德國(guó)數(shù)學(xué)家高斯是近代數(shù)學(xué)奠基者之一,有“數(shù)學(xué)王子”之稱,在歷史上有很大的影響.他幼年時(shí)就表現(xiàn)出超人的數(shù)學(xué)天賦,10歲時(shí),他在進(jìn)行1+2+3+…+100的求和運(yùn)算時(shí),就提出了倒序相加法的原理,該原理基于所給數(shù)據(jù)前后對(duì)應(yīng)項(xiàng)的和呈現(xiàn)一定的規(guī)律生成,因此,此方法也稱為高斯算法.已知某數(shù)列通項(xiàng)an=2n?1002n?101,則a1